While an equation might be true for some specific value of variables within it, a mathematical identity is true for every value of the variables within it. Many of the patterns used in factoring polynomials are mathematical identities.

Exploration

Press the 'Play/Pause' button to progress through the slides. Or use the slider to change the slides more slowly.

Use your observations to answer the questions.

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In slide 1, what is the area of the red square in terms of a?
In slide 2, what is the area of the blue square in terms of b?
If the blue square is taken away, what expression can we use to represent the area that remains?
In slides 3 and 4, what's left of the square is rearranged to create a rectangle. Write the dimensions of the new rectangle in terms of a and b.
Write the area of the new rectangle in terms of a and b.
Explain how the equation you found in question 3 is equal to the area of the rectangle you found in question 5.

One mathematical identity we have worked with already is the difference of squares, m^{2}-n^{2}=\left(m-n\right)\left(m+n\right). It is easily proven using a geometric diagram such as the one from the exploration.

Another familiar identity we have previously used is {a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2}} when factoring perfect square trinomials. Two more important identities are the sum of cubes and difference of cubes.

Sum of cubes identity

Two perfect cube expressions being added to each other

a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)

Difference of cubes identity

Two perfect cube expressions being subtracted from each other

a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

In the proof of an identity, it is our job to prove both sides are equal, so we must work with one side of the equation and show that algebraic manipulation leads to the other side. We cannot manipulate both sides of the equation at once because changing both sides assumes that both sides are already equal.

Examples

Example 1

The perfect square trinomial identity is \left(a+b\right)^{2}=a^{2}+2ab+b^{2}.

Prove the identity.

Worked Solution

Create a strategy

We can prove the identity using a geometric diagram.

Apply the idea

We will begin by drawing a square with side lengths of a+b. The area of this square is the left side of the equation, \left(a+b\right)^{2}.

A square divided into 2 rows of 2 rectangles. The outside of the square is labeled a above the left column and b above the right column. The side is labeled a next to the top row and b next to the bottom row.

Next, we can find the areas of the squares and rectangles within the larger square.

A square divided into 2 rows of 2 rectangles. From top to bottom, the left column of squares is labeled a squared and a b, and the right column of squares is labeled a b and b squaed. The outside of the square is labeled a above the left column and b above the right column. The side is labeled a next to the top row and b next to the bottom row.

The area of the larger square is equal to the sum of the areas of the squares and rectangles within it. Therefore, \left(a+b\right)^{2}=a^{2}+2ab+b^{2}.

Reflect and check

Alternatively, we could prove the identity through algebraic manipulation. We are trying to prove \left(a+b\right)^{2}=a^{2}+2ab+b^{2} We can only manipulate one side of the equation, so we will expand the left hand side using the distributive property.

\displaystyle \left(a+b\right)^{2}	\displaystyle =	\displaystyle \left(a+b\right)\left(a+b\right)	Expand the power
	\displaystyle =	\displaystyle a^{2}+ab+ba+b^{2}	Distributive property
	\displaystyle =	\displaystyle a^{2}+ab+ab+b^{2}	Commutative property of multiplication
	\displaystyle =	\displaystyle a^{2}+2ab+b^{2}	Add like terms

Use the identity to evaluate 98^{2}.

Worked Solution

Create a strategy

Because 98 is squared, we can represent it with the left side of the equation.\left(a+b\right)^{2}=98^{2}

Apply the idea

When choosing values for a and b, we want to choose values that would be easy to calculate with. One option is to use a=100 and b=-2. \left(100+-2\right)^{2}=98^{2}

Now, we can use the right side of the identity to easily calculate the value of 98^{2}.

\displaystyle \left(a+b\right)^{2}	\displaystyle =	\displaystyle a^{2}+2ab+b^{2}	State the identity
\displaystyle \left(100+-2\right)^{2}	\displaystyle =	\displaystyle \left(100\right)^{2}+2\left(100\right)\left(-2\right)+\left(-2\right)^{2}	Substitute a=100 and b=-2
\displaystyle 98^{2}	\displaystyle =	\displaystyle 10\,000-400+4	Evaluate the multiplication and powers
\displaystyle 98^{2}	\displaystyle =	\displaystyle 9604	Evaluate the addition and subtraction

Reflect and check

You can use your calculator to confirm this answer, but this is a helpful strategy to use when calculator use is not permitted.

Example 2

Use an identity to expand the expression \left(x - 6\right) \left(x^{2} + 6 x + 36\right). State which identity was used.

Worked Solution

Create a strategy

When we multiply the first terms in both sets of parentheses, we get an x^{3} term. This tells us that the expression must use one of the cube identities, sum of cubes or difference of cubes.

Apply the idea

The given expression \left(x - 6\right) \left(x^{2} + 6 x + 36\right) represents the difference of cubes identity, as it follows the pattern a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + ab + b^{2}\right).

In this case, a=x and b=6, so:

\displaystyle \left(x - 6\right) \left(x^{2} + 6 x + 36\right)	\displaystyle =	\displaystyle x^{3}-6^{3}
	\displaystyle =	\displaystyle x^{3}-216

Reflect and check

We can check our result by expanding the expression using the distributive property. \begin{aligned}\left(x - 6\right) \left(x^{2} + 6 x + 36\right)&=x\left(x^{2}\right)+x\left(6x\right)+x\left(36\right) -6\left(x^{2}\right)-6\left(6x\right)-6\left(36\right)\\&=x^{3}+6x^{2}+36-6x^{2}-36x-216\\&=x^{3}\cancel{+6x^{2}-6x^{2}}\cancel{+36x-36x}-216\\&=x^{3}-216\end{aligned}Notice that this method requires more steps, so it is more efficient to expand by using a polynomial identity.

Example 3

Factor the following expressions and identify which identity was used.

25r^{2} - 60rs + 36s^{2}

Worked Solution

Create a strategy

Look for patterns that match known polynomial identities to factor the expression. Notice that the first and last terms, 25 r^{2} and 36 s^{2}, are perfect squares. They can be rewritten as \left(5r\right)^{2} and \left(6s\right)^{2}.

Also, observe that the middle term, -60 r s, is twice the product of the square roots of the first and last terms.-2\left(5r\right)\left(6s\right)=-60rs This suggests we can factor the expression using the perfect square trinomial identity \left(a-b\right)^{2} = a^{2}-2ab+b^{2}.

Apply the idea

The expression 25r^{2} - 60rs + 36s^{2} can be factored using the identity a^{2}-2ab+b^{2}=\left(a-b\right)^{2} where a=5r and b=6s.

\displaystyle 25r^{2} - 60rs + 36s^{2}

\displaystyle =

\displaystyle \left(5r-6s\right)^{2}

Reflect and check

One way to check our result is by substituting a specific value for r and s into both the original and factored expressions and verifying that they yield the same result.

Let's check for r=1 and s=2:

\displaystyle 25\left(1\right)^{2} - 60\left(1\right)\left(2\right) + 36\left(2\right)^{2}	\displaystyle =	\displaystyle \left(5\left(1\right)-6\left(2\right)\right)^{2}	Substitute r = 1 and s=2 into both expressions
\displaystyle 25-120+144	\displaystyle =	\displaystyle \left(5-12\right)^{2}	Evaluate the multiplication
\displaystyle 49	\displaystyle =	\displaystyle 49\quad \checkmark	Simplify both sides

64m^{6} - 81n^{4}

Worked Solution

Create a strategy

This expression has two terms that are subtracted. The only identity that has the same structure is the difference of squares identity. Although it may not be obvious at first, both terms are in fact perfect squares, so we can use the difference of squares identity to factor the expression.

Apply the idea

The expression is an example of the difference of squares, following the pattern a^{2}-b^{2}=\left(a+b\right)\left(a-b\right).

We can rewrite both terms as perfect squares:

64m^6=\left(8m^3\right)^2
81n^4=\left(9n^2\right)^2

This shows a=8m^{3} and b=9n^{2}.

\displaystyle 64m^{6} - 81n^{4}	\displaystyle =	\displaystyle \left(8m^{3}\right)^{2}-\left(9n^{2}\right)^{2}
	\displaystyle =	\displaystyle \left(8m^{3}+9n^{2}\right)\left(8m^{3}-9n^{2}\right)

Reflect and check

Although this expression was the difference of two square terms, not every binomial with a subtraction sign represents the difference of two squares. For example, 50x^{2}-9 is not a difference of two perfect squares.

We have to be careful that we only apply the identity when the terms are in fact perfect squares because the identity cannot be used on non-square terms.

125g^{3}+64h^{3}

Worked Solution

Create a strategy

The given expression is a sum of two terms, both of which are perfect cubes. This means we can use the sum of cubes identity to factor the expression.

Apply the idea

The expression is an example of the difference of cubes identity, a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

When can rewrite both terms as perfect cubes:

125g^{3}=\left(5g\right)^{3}
64h^{3}=\left(4h\right)^{3}

This shows a=5g and b=4h.

\displaystyle 125g^{3}+64h^{3}	\displaystyle =	\displaystyle \left(5g+4h\right)\left(\left(5g\right)^{2}-\left(5g\right)\left(4h\right)+\left(4h\right)^{2}\right)
	\displaystyle =	\displaystyle \left(5g+4h\right)\left(25g^{2}-20gh+16h^{2}\right)

Reflect and check

The factorization for the sum of cubes and difference of cubes are very similar. The only difference between them is the signs of the terms.\begin{aligned}a^{3}+b^{3}&=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)\\a^{3}-b^{3}&=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)\end{aligned}To remember the difference between the signs of each identity, we can use the acronym SOAP:

S (Same): The first sign (in the binomial factor) is the SAME as the sign in the original expression. This reflects the "S" in SOAP.
O (Opposite): The second sign (first sign in the trinomial factor) is OPPOSITE to the original expression's sign, aligning with the "O" in SOAP.
AP (Always Positive): The last sign (second sign in the trinomial factor) is ALWAYS POSITIVE, regardless of the original expression's sign. This corresponds to the "AP" in SOAP.

Example 4

Use the algebra tiles given to verify that x^{2}-6x+9=\left(x-3\right)^{2}.

Algebra tiles: one positive square with x by x measurement label, then six negative rectangles with x by 1 measurement label, and nine positive small squares with 1 by 1 measurement label.

Worked Solution

Create a strategy

The first algebra tile has an area of x^{2}, representing the first term in the trinomial. The next group of 6 tiles each have an area of x, but their sign is negative. This represents the second term, -6x. The remaining group of 9 square tiles each have an area of 1, representing the constant term 9.

To verify this identity, we can rearrange the algebra tiles to form a square with a side length of x-3.

Apply the idea

Algebra tiles: 1 positive square tile, 6 negative rectangle tiles and 6 positive small square tiles. Arranged to create a square, then sides are labeled with 'x-3'.

This verifies that x^{2}-6x+9=\left(x-3\right)^{2}.

Reflect and check

Let's expand the expression \left(x-3\right)^{2} using the distributive property to check our answer algebraically.

\displaystyle \left(x-3\right)^{2}	\displaystyle =	\displaystyle \left(x-3\right)\left(x-3\right)
	\displaystyle =	\displaystyle x\left(x-3\right)-3\left(x-3\right)
	\displaystyle =	\displaystyle x^{2}-3x-3x+9
	\displaystyle =	\displaystyle x^{2}-6x+9

Idea summary

Important identities we use often are:

Perfect square trinomials: a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}
Difference of squares: a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)
Sum of cubes: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)
Difference of cubes: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

We can verify identities mathematically through algebraic manipulation or using geometric diagrams. These identities can be used to describe numerical relationships.

Outcomes

A2.EO.3

The student will perform operations on polynomial expressions in two or more variables and factor polynomial expressions in one and two variables.

A2.EO.3a

Determine sums, differences, and products of polynomials in one and two variables.

A2.EO.3b

Factor polynomials completely in one and two variables with no more than four terms over the set of integers.

A2.EO.3d

Represent and demonstrate equality of polynomial expressions written in different forms and verify polynomial identities including the difference of squares, sum and difference of cubes, and perfect square trinomials.

3.02 Polynomial identities

Ideas

Polynomial identities

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

A2.EO.3

A2.EO.3a

A2.EO.3b

A2.EO.3d

What is Mathspace

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